Both scenarios are possible?

$1+5=6$ and $1\times5=5$

$2+3=5$ and $2\times3=6$

$5$ and $6$ are not divisible by $4$ , so $4$ cannot be a value for the 2 numbers.

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Solving for non-integer solutions, with variables $x$ and $y$ ( $x>y$ ):

We have considered the cases where $x+y=6$ , $x\times y=5$ and $x+y=5$ , $x\times y=6$ .

If $x+y=6$ and $x\times y=6$ , then $y=\frac{6}{x}$ and $x+\frac{6}{x}=6$ .

$\implies x^2-6x+6=0 \implies x=\boxed{3+\sqrt{3}} \implies y=\boxed{3-\sqrt{3}}$ .

Or if $x+y=5$ and $x\times y=5$ , then $y=\frac{5}{x}$ and $x+\frac{5}{x}=5$ .

$\implies x^2-5x+5=0 \implies x=\boxed{\frac{5+\sqrt{5}}{2}} \implies y=\boxed{\frac{5-\sqrt{5}}{2}}$ .

Clearly $\left(2,3\right)$ and $\left(1,5\right)$ are solutions.

Neither of the numbers can be $4$ , since if either were $4$ then the other one wouldn't be an integer since neither $5$ nor $6$ are divisible by $4$ , which would be impossible since sum of integer and non-integer cannot be an integer.

Hence, the answer is $\boxed{4}$ .

$(1, 5)$ and $(2, 3)$ are solutions, so they can't be the answer. We will show that $4$ cannot be either of the numbers.

If we assume the first number is $4$ , we have $4 + y = 5$ or $4 + y = 6$ . This would yield either $y = 1$ or $y = 2$ , respectively. But neither of these satisfy $4y \in \{5, 6\}$ . as $4 \cdot 1 = 4$ and $4 \cdot 2 = 8$ . Thus $4$ can't be a value for either of the two numbers.

Because of the multiple choice the question becomes easy: 1 and 5 together fulfill the requirements and so do 3 an 2. So 4 is the answer to this question. Next to that: 4 doesn't divide 5 nor 6, so it can't be 1 of the numbers either.

Neither 5 nor 6 are divisible by 4, so multiplication would be an utter disaster. This basically gives away the answer.